# Motion of sleeve attached to a L-shaped rotating rod

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1. Jul 20, 2017

### Pushoam

1. The problem statement, all variables and given/known data
View attachment 207518

View attachment 207517
View attachment 207519
2. Relevant equations

3. The attempt at a solution
Let me assume that initially the spring is at its natural length.

The normal force acting in the vertical direction will manage the gravitational force.
And the normal force acting in the radially inward direction will provide the centripetal acceleration.
There is no force acting along the line AB. So, why will the sleeve move along AB at the first place?

Last edited: Jul 20, 2017
2. Jul 20, 2017

### .Scott

You are right. If the initial condition is spring length = 0 and the length of the sleeve = 0, then it will stay that way.
If I was assigned this, I would also check to see if there was any other equilibrium states for any combinations of length OB, omega, x, and m.
There is also part 2 the question. How does the direction of rotation affect the reponse?

3. Jul 20, 2017

### Pushoam

How could length of spring and sleeve be zero?
I am not getting why the sleeve moves along AB.

4. Jul 20, 2017

### .Scott

How can the mass of the spring be zero?
If the center of mass of the sleeve is not at B, then the angle between the centrifugal force and AB will not be 90 degrees - so a component of the centrifugal force will pull the mass away from B and stretch the spring. The further the sleeve gets from B, the greater the component of the centrifugal force will be applied in the B->A direction.

5. Jul 20, 2017

### Pushoam

Can anyone please tell me how to make the image visible in the OP instead of appearing as an attachment?
Because the image consists of the question.

6. Jul 20, 2017

### .Scott

Use "img" and "/img" in square brackets:

7. Jul 20, 2017

### haruspex

That assumes the relaxed length is zero.
It seems to me that the question has omitted to specify a relaxed length. Let it be L.
If L>0 then the normal force is not radial to the axis. The spring must provide a force so that the resultant is radial.

8. Jul 20, 2017

### Pushoam

In this quote,I am talking about the forces acting on the sleeve.
In the above post, what I meant is:
Let's say that the natural length of the spring is l0 >0. So, the sleeve is at a distance l0 from B, initially.
Now, using the argument given in the above post I claimed that the sleeve does not move along AB as there is no force acting along this direction.

But,what I understand from your statements is that
there will be a force called centrifugal force acting on the sleeve along AB which will create its motion along AB.
The effect of this centrifugal force will tend to zero if we approximate l0 to 0 i.e.if the sleeve is at B and no spring force is acting on it.

Is my understanding correct so far?

9. Jul 21, 2017

### haruspex

That is amost correct, using the non-inertial frame of the sleeve. To clarify, though, the centrifugal force would act along OA and so have a component along BA.

In an inertial frame, you can say that in order to stay at the same point on the rod the net force on the sleeve would have to along AO, i.e. purely centripetal. With the spring relaxed, the only forces available are normal to the rod. Thus the system cannot be stable and the sleeve must move relative to the rod.

10. Jul 24, 2017

### Pushoam

The surface of the rod is cylindrical. So, there can be a component of contact force (which is normal to the surface, not friction) along AO. As contact force is self - adjustable, we can't say that the sleeve must move relative to the rod, can we?

However,since the question mentions that the sleeve moves relative to the rod, we take this component negligible.

Is this correct?

11. Jul 24, 2017

### haruspex

I take L-shaped to mean OBA is a right angle. The normal force at A lies in a plane normal to BA, and that does not include O.

12. Jul 24, 2017

### CWatters

With the arm stationary the spring will pull the sleeve to a position corresponding to its natural length. When rotating the sleeve will move to a different position along the rod. At that position..

What do you know about the direction of the net force acting on the sleeve?
What two forces contribute to that net force and which directions do they act?

13. Jul 25, 2017

### Pushoam

The surface in contact is cylindrical. So, the normal force will be in the $\hat s$ . At some angle $\phi$ , the normal force could be along AO, couldn't it?

Wrt rod frame,
Let's say that at time t, the sleeve is moving towards B from A and the spring is elongated with elongation le. The direction from A to B is defined to be $\hat l$. l is the distance of the sleeve from A where A is the other end of the rod.

velocity of the sleeve = $\vec v_r = \dot l \hat l$
acceleration of the sleeve = $\vec a_r = \ddot l \hat l$

Now, eqn. of motion is given as

$\vec F_n = \vec F_p + \vec F_f$
where $\vec F_n$ = net fofrce acting on the sleeve wrt the rod frame
$\vec F_p$ = net physical force acting on the system =$k l_e \hat l$ i.e.spring force + $\vec f$ i.e. friction + $\vec N$ i,e normal force + m $\vec g$ where k is the spring constant
$\vec F_f$= net fictitious force acting on the system = $m\omega^2 r~ \hat r - 2m\omega ~\hat z \times \dot l~\hat l$

$m\ddot l \hat l = k l_e \hat l - f \hat l +\vec N + m\vec g + m\omega^2 r~ \hat r - 2m\omega ~\hat z \times \dot l~\hat l$
Taking $\vec N + m\vec g =0$
$\ddot l \hat l= \{kl_e - \frac f m \}\hat l+ \omega^2 r~ \hat r - 2\omega ~\hat z \times \dot l~\hat l$

Note that $\dot l = \frac {dl} {dt}$ looks as big form of small i.
Is this correct so far?

Now,

$\ddot l = - \ddot l_e$
$\dot l = - \dot l_e$
$l= AB -\{ l_0 + l_e\}$
$r = \sqrt { \{l_0 + l_e \}^2 + {OB}^2 }$
$\hat z \times \hat l = \hat m$ where $\hat m$ is a unit vector along BO
$\hat r = \cos \theta \hat m +\sin \theta \hat l$ where $\theta$ is the angle which $\hat r$ makes with BO.
Taking friction to be 0,
$\ddot l \hat l= \{kl_e - \frac f m \}\hat l+ \omega^2 r~ \hat r - 2\omega ~\hat z \times \dot l~\hat l$
$- \ddot l_e \hat l= kl_e\hat l+ \omega^2 \sqrt { \{l_0 + l_e \}^2 + {OB}^2 }~\{\cos \theta \hat m +\sin \theta \hat l \} + 2\omega \dot l_e ~\hat m$

we have the following two equations,
$- \ddot l_e = kl_e+ \omega^2 \sqrt { \{l_0 + l_e \}^2 + {OB}^2 }\sin \theta$
$0= \omega^2 \sqrt { \{l_0 + l_e \}^2 + {OB}^2 }~\cos \theta + 2\omega \dot l_e$

Is it correct so far?

Last edited: Jul 25, 2017
14. Jul 25, 2017

### haruspex

The normal to a cylindrical surface is at right angles to the axis of the cylinder, i.e. at right angles to AB. OB is also at right angles to AB, and the length AB is nonzero. There is no way that the normal can pass through O.

15. Jul 25, 2017

### Nidum

@Pushoam The sketch given in the problem statement is not very good and may be misleading you . This is the problem geometry shown as a true view looking vertically downwards :

16. Jul 25, 2017

### Pushoam

Thanks a lot for this sketch.

17. Jul 27, 2017

### Pushoam

Let's denote the direction OB by $\hat r$ and BA by $\theta$.
Wrt rod frame,
Let's say that at time t,
the sleeve is moving towards B from A and the spring is elongated with elongation $l_e$.
l denotes the distance of the sleeve from B
the position of sleeve is given by R\hat R = $r\hat r + l \theta$ where r = OB
elongation$l_e = l- l_0$ , where l0 is the natural length of the spring
the velocity of the sleeve $\vec v = \frac {d \{R\hat R\}} {dt} = \dot l \theta = \dot l_e\hat \theta$
the acceleration of the sleeve $\vec a = \ddot l_e \hat\theta$

Now, eqn. of motion is given as

$\vec F_n = \vec F_p + \vec F_f$
where $\vec F_n$ = net force acting on the sleeve wrt the rod frame
$\vec F_p$= net physical force acting on the system =$- k l_e \hat \theta$ i.e.spring force + ⃗ $\vec f$ i.e. friction + ⃗$\vec N$ i,e normal force + m ⃗$\vec g$ where k is the spring constant
⃗$\vec F_f$ = net fictitious force acting on the system =$m\omega^2 R~ \hat R - 2m\omega ~\hat z \times \dot l_e~\hat\theta$

$m\ddot l_e \hat \theta =- k l_e \hat\theta - f \hat\theta +\vec N + m\vec g + m\omega^2R~ \hat R - 2m\omega ~\hat z \times \dot l_e~\hat \theta$
Taking $\vec N + m\vec g =0$
$\ddot l_e \hat \theta = \{- kl_e - \frac f m \}\hat\theta+ \omega^2 R~ \hat R - 2\omega ~\hat z \times \dot l_e~\hat \theta$

Is this correct so far?

18. Jul 27, 2017

### haruspex

No, the normal force also has a horizontal component normal to the rod.
For the rest of it, it looks ok, but I am never comfortable working with non-inertial frames.

19. Jul 29, 2017

### Pushoam

Taking $\vec N + m\vec g = N\hat r - mg \cos \alpha \hat r +mg\sin \alpha \hat \theta$
$m\ddot l_e \hat \theta =- k l_e \hat\theta - f \hat\theta +\vec N + m\vec g + m\omega^2R~ \hat R - 2m\omega ~\hat z \times \dot l_e~\hat \theta$
$m\ddot l_e \hat \theta =- k l_e \hat\theta - f \hat\theta +N\hat r - mg \cos \alpha \hat r +mg\sin \alpha \hat \theta + m\omega^2R~ \hat R - 2m\omega ~\hat z \times \dot l_e~\hat \theta \\ \text { ignoring } m\omega^2R~ \hat R ~ and ~f \hat\theta~, \\ \ddot l_e =\frac {k l_e} m + g\sin \alpha \\ l_e = A\cos \sqrt {\frac k m } t + B \sin \sqrt {\frac k m } t + \frac{ m g \sin \alpha}{k}$

But this is not the required answer as the answer has to depend on $\omega$ and I ignored it. Am I on the right track?

20. Jul 29, 2017

### Nidum

Try to formulate your solutions in terms of just one variable - this can be angle AOB or radius R or the stretched/compressed length AB of the spring .

I've shown the sleeve as a point mass in the lower diagram - what are the forces acting on it ?

(Two cases - one for each direction of rotation of the swing arm)

nb: Your angle alpha is not needed to solve this problem .

Last edited: Jul 29, 2017